Pricing a Put with a Future-Determined Strike in Black–Scholes
Summary
The contract pays the positive part of the difference between the stock price at an earlier date and its price at maturity. At the earlier date, the strike is known, so the claim becomes an ordinary European put with a strike equal to that observed stock price. Under Black–Scholes assumptions, its value at that date is proportional to the stock price, with a proportionality factor determined by rates, dividends, volatility, and the remaining term.
For valuation before the strike-setting date, the solution uses conditional expectation in two stages: first value the put at the date its strike becomes known, then discount and take the risk-neutral expectation back to the earlier valuation time. The stock’s risk-neutral expected growth yields the pre-date price. The derivation assumes the Black–Scholes framework and its stated market parameters; it does not address model risk or alternative dynamics.
Key ideas
- At the strike-setting date, the payoff is that of a standard put struck at the then-current stock price.
- The put value on that date is proportional to the stock price under Black–Scholes assumptions.
- Before that date, condition first on the strike-setting date and then take the risk-neutral expectation backward.
- The resulting value depends on the model’s rates, dividend yield, volatility, and time intervals.
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Full text
# Pricing of $(S(T_0)-S(T))^+$
# Pricing of $(S(T_0)-S(T))^+$
> Problem: Consider a new derivative that at time $T$ pays $Y =(S(T_0) − S(T))^+$
> where $0 < T_0 < T$ is a fixed date.
> (i) Show that the arbitrage-free of Y at time $t = T_0$ is given by $\pi_{T_0} (Y ) =pS(T_0)$ where p is independent of the stock price.
> (ii) Determine the arbitrage-free price of derivative $Y$ at time $t < T_0$.
To value the contract I would use the Black-Scholes PDE solution $E^{Q}((S(T_0) − S(T))^+|\mathscr{F}_{T_{0}})=E^{Q}((S(T_0) − S(T)) \:1_{S(T_0\geqslant S(T))}|\mathscr{F}_{T_{0}})$
I know that $S(T_0)-S(T)$ is independent of $\mathscr{F}_{T_0}$. But it is not independent of $1_{S(T_0\geqslant S(T))}$, so I cannot split them into two expected values. Even if I did, I would not get the result. I tried to solve using the exponential form of S(T) but got nowhere.
I though that when $t=T_0$ then $Y$ is a put option which implies that $E^{Q}((S(T_0) − S(T))^+|\mathscr{F}_{T_{0}})=S(T_0)(e^{T-T_0}\Phi(-d_2)-\Phi(-d_1))$ where d_2 and d_1 do not depend on the stock.
Question:
Is my solution correct?
How do I solve the second question when $t<T_0$? I think I cannot used a pre-defined result.
Thanks in advance!
## Answer by Kevin (score 4, accepted)
https://quant.stackexchange.com/a/61282
At time $T_0$, the strike price becomes known and the option turns into a ``normal'' put option, i.e. \begin{align*} V(T_0,S_{T_0}) &= S_{T_0}e^{-r(T-T_0)}\Phi(-d_2)-S_{T_0}e^{-q(T-T_0)}\Phi(-d_1) \\ &= S_{T_0}\underbrace{\left(e^{-r(T-T_0)}\Phi(-d_2)-e^{-q(T-T_0)}\Phi(-d_1)\right),}_{=:p} \end{align*} where $p$ is indeed independent of the stock price because \begin{align*} d_{1,2}=\frac{r-q\pm\frac{1}{2}\sigma^2}{\sigma}\sqrt{T-T_0}. \end{align*}
In general, for $t> T_0$, the Black-Scholes formula is \begin{align*} V(t,S_t) &= S_{T_0}e^{-r(T-t)}\Phi(-d_2)-S_te^{-q(T-t)}\Phi(-d_1), \end{align*} where \begin{align*} d_{1,2}=\frac{\ln\left(\frac{S_t}{S_{T_0}}\right)+\left(r-q\pm\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}}. \end{align*} Of course, $\lim\limits_{t\downarrow T_0}V(t,S_t)=S_{T_0}p$.
For $t<T_0$, the option value is \begin{align*} V(t,S_t)&=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\max\{S_{T_0}-S_T,0\}\right] \\ &=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\mathbb{E}^\mathbb{Q}_{T_0}\left[\max\{S_{T_0}-S_T,0\}\right]\right] \\ &=e^{-r(T_0-t)}\mathbb{E}^\mathbb{Q}_t\left[e^{-r(T-T_0)}\mathbb{E}^\mathbb{Q}_{T_0}\left[\max\{S_{T_0}-S_T,0\}\right]\right]\\ &=e^{-r(T_0-t)}\mathbb{E}^\mathbb{Q}_t\left[S_{T_0}p\right] \\ &=pe^{-r(T_0-t)}S_te^{(r-q)(T_0-t)} \\ &=S_te^{-q(T_0-t)}\left(e^{-r(T-T_0)}\Phi(-d_2)-e^{-q(T-T_0)}\Phi(-d_1)\right). \end{align*} Of course, $\lim\limits_{t\uparrow T_0}V(t,S_t)=S_{T_0}p$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.